Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline int Gcd(int X, int Y) { return Y ? Gcd(Y, X % Y) : X; }
inline long long Gcd(long long X, long long Y) { return Y ? Gcd(Y, X % Y) : X; }
inline int Pow(int base, long long exp, int _mod) {
  if (!(base %= _mod)) return 0;
  int _ans = 1;
  for (; exp; exp >>= 1, base = (long long)base * base % _mod)
    exp& 1 ? _ans = (long long)_ans * base % _mod : 0;
  return _ans;
}
inline long long Pow(long long base, long long exp, long long _mod) {
  if (!(base %= _mod)) return 0;
  long long _ans = 1;
  for (; exp; exp >>= 1, base = base * base % _mod)
    exp& 1 ? _ans = _ans * base % _mod : 0;
  return _ans;
}
const int INF = 0x3f3f3f3f;
int sy, a, b, n, q;
int C[400005], S[400005];
double c[400005];
int main() {
  scanf("%d%d%d%d", &sy, &a, &b, &n);
  for (int i = 1; i <= (n); ++i)
    scanf("%d%d", C + i + i - 1, C + i + i),
        c[i + i - 1] = C[i + i - 1], c[i + i] = C[i + i],
                  S[i + i - 1] = S[i + i - 2],
                  S[i + i] = S[i + i - 1] + C[i + i] - C[i + i - 1];
  scanf("%d", &q);
  for (int i = 1; i <= (q); ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    double xl = a + (double)(x - a) * (-sy) / (y - sy);
    double xr = b + (double)(x - b) * (-sy) / (y - sy);
    int pl = lower_bound(c + 1, c + n + n + 1, xl) - c;
    int pr = upper_bound(c + 1, c + n + n + 1, xr) - c - 1;
    if (pr < pl) {
      printf("%.15lf\n", (double)(pl & 1 ? 0 : b - a));
      continue;
    }
    double dl = xr - xl, s = S[pr] - S[pl];
    if (~pl & 1) s += C[pl] - xl;
    if (pr & 1) s += xr - C[pr];
    printf("%.15lf\n", s * (b - a) / dl);
  }
  return 0;
}